Question

Ajouin and Genie are a happy couple and have 4 lovely children. Is it more likely they will have two boys and two girls or three of the same sex and one of the other? Assume that the probability of a child being a boy is 0.5 and that the births are independent events. (5 points)

2.    A fair die is rolled 4 times. Let the random variable X denote the number of sixes that appear. Find F_X(x). (5 points)

Answer 1

1.

$$P(bbbb)=(\dfrac{1}{2})^4=\dfrac{1}{16}$$

$$P(bbbg)=P(bbgb)=P(bgbb)=P(gbbb)$$

$$=(\dfrac{1}{2})^4=\dfrac{1}{16}$$

$$P(bbgg)=P(bgbg)=P(bggb)=P(gbbg)=P(gbgb)=P(ggbb)$$

$$=(\dfrac{1}{2})^4=\dfrac{1}{16}$$

$$P(gggb)=P(ggbg)=P(gbgg)=P(bggg)$$

$$=(\dfrac{1}{2})^4=\dfrac{1}{16}$$

$$P(gggg)=(\dfrac{1}{2})^4=\dfrac{1}{16}$$

$$P(2\ boys\ \&\ 2\ girls)=6(\dfrac{1}{16})=\dfrac{3}{8}$$

$$P(3\ boys \ \&\ 1\ girl)+P(1\ boy \ \&\ 3\ girls)$$

$$=4(\dfrac{1}{16})+4(\dfrac{1}{16})=\dfrac{1}{2}>\dfrac{3}{8}$$

It is more likely they will have three of the same sex and one of the other than two boys and two girls.

2. $X\sim Bin(n, p)$

$n=4, p=\dfrac{1}{6}, q=1-p=\dfrac{5}{6}$

$$P(X=0)=\dbinom{4}{0}(\dfrac{1}{6})^0(\dfrac{5}{6})^{4-0}=\dfrac{625}{1296}$$

$$P(X=1)=\dbinom{4}{1}(\dfrac{1}{6})^1(\dfrac{5}{6})^{4-1}=\dfrac{500}{1296}$$

$$P(X=2)=\dbinom{4}{2}(\dfrac{1}{6})^2(\dfrac{5}{6})^{4-2}=\dfrac{150}{1296}$$

$$P(X=3)=\dbinom{4}{3}(\dfrac{1}{6})^3(\dfrac{5}{6})^{4-3}=\dfrac{20}{1296}$$

$$P(X=4)=\dbinom{4}{4}(\dfrac{1}{6})^4(\dfrac{5}{6})^{4-4}=\dfrac{1}{1296}$$

$$F_X(x) = \begin{cases} 0, &x<0 \\ 625/1296, &0\leq x<1\\ 1125/1296, &1\leq x<2\\ 1275/1296, &2\leq x<3\\ 1295/1296, &3\leq x<4\\ 1, &x\geq 4. \end{cases}$$